Covariance in Bivariate Normal Models

Why Covariance Matters in a Bivariate Normal Model

Two variables can each have a familiar normal distribution while their joint behavior still depends on how they move together. For a bivariate normal model, that dependence is summarized by one parameter: either the covariance or, equivalently, the correlation. Understanding how that parameter enters the density lets you interpret the model, check whether its parameters are possible, and decide when zero correlation does imply independence.

Throughout, consider two continuous random variables \(X\) and \(Y\) with means \(\mu_X,\mu_Y\), positive standard deviations \(\sigma_X,\sigma_Y\), and correlation \(\rho\). The ordinary density formula below applies when \(-1<\rho<1\). The endpoints \(\rho=-1\) and \(\rho=1\) describe singular cases, not a two-dimensional density of this form.

The covariance is sometimes used directly as the dependence parameter. Writing \(c=\operatorname{Cov}(X,Y)\), the same relationship becomes \(\rho=c/(\sigma_X\sigma_Y)\). Thus the two parameters measure dependence on different scales: covariance has units equal to the product of the variables’ units, whereas correlation is dimensionless and lies between \(-1\) and \(1\).

The density’s cross-product term, \(-2\rho z_Xz_Y\), where \(z_X=(x-\mu_X)/\sigma_X\) and \(z_Y=(y-\mu_Y)/\sigma_Y\), is the part that reflects dependence. If \(\rho\) is positive, points for which both standardized values have the same sign are favored relative to points with opposite signs. If \(\rho\) is negative, the opposite-sign combinations are favored. The overall density remains a joint model; the cross-product term should not be interpreted in isolation from the other terms and the normalizing constant.

Why the Covariance Parameter Is \(\rho\sigma_X\sigma_Y\)

A useful way to verify the covariance relationship is to express the pair using two independent standard normal variables \(Z_1\) and \(Z_2\). For \(-1<\rho<1\), write

Both standardized components have mean zero. Since \(Z_1\) and \(Z_2\) are independent with variance one, the variance of the standardized component of \(Y\) is \(\rho^2+(1-\rho^2)=1\). The covariance follows by expanding the product of the centered variables:

Here \(\operatorname{Cov}(Z_1,Z_2)=0\) because the standard normal variables are independent, and \(\operatorname{Var}(Z_1)=1\). This construction also explains the density: transforming \((Z_1,Z_2)\) to \((X,Y)\) gives the stated bivariate normal density by the change-of-variables formula. In particular, the absolute Jacobian determinant is \(\sigma_X\sigma_Y\sqrt{1-\rho^2}\), which accounts for the density’s normalizing factor.

Reading the Density and Checking the Parameters

The covariance and correlation cannot be chosen independently of the marginal standard deviations. Since \(\rho=c/(\sigma_X\sigma_Y)\), a proposed covariance \(c\) must satisfy \(|c|<\sigma_X\sigma_Y\) for the nondegenerate density above. Equivalently, the covariance matrix

must have positive determinant: \(\sigma_X^2\sigma_Y^2-c^2>0\). If equality holds, then \(|\rho|=1\), and the joint distribution lies on a straight line rather than having the displayed two-dimensional density. If \(|c|>\sigma_X\sigma_Y\), the proposed parameters are invalid.

When using the density, first calculate the standardized values \(z_X\) and \(z_Y\), then substitute them into the entire quadratic expression. In particular, preserve the signs of both standardized values in the cross-product. A negative \(z_Y\) changes the sign of \(z_Xz_Y\), which can change the exponent substantially.

Zero Correlation and Independence

Zero covariance does not generally establish independence. In the nondegenerate bivariate normal model, however, the zero-correlation case has a special property: the joint density factors into the product of the marginal densities. This factorization is the direct reason that \(X\) and \(Y\) are independent when \(\rho=0\).

Set \(\rho=0\) in the joint density. The cross-product term disappears, the denominator contains \(\sqrt{1-\rho^2}=1\), and the exponent separates into a term involving only \(x\) and a term involving only \(y\):

A joint density that equals the product of its marginal densities gives independence: probabilities over rectangular regions factor into the corresponding marginal probabilities. Conversely, if \(X\) and \(Y\) are independent and have finite variances, then their covariance is zero. Because the standard deviations in this model are positive, \(\operatorname{Cov}(X,Y)=\rho\sigma_X\sigma_Y=0\) implies \(\rho=0\). Thus, for a nondegenerate bivariate normal pair, independence and zero correlation are equivalent.

Common Exam Traps and Practical Checks

• Confusing covariance with correlation. The correct conversion is \(\rho=c/(\sigma_X\sigma_Y)\). For example, if the standard deviations are \(4\) and \(10\) and the covariance is \(24\), then \(\rho=24/(4\cdot10)=0.6\), not \(24\).
• Ignoring the allowed parameter range. A nondegenerate density requires \(|c|<\sigma_X\sigma_Y\). A proposed covariance of \(45\) with standard deviations \(5\) and \(8\) is impossible for this model because \(45>5(8)=40\).
• Assuming zero correlation always means independence. That implication is valid for the bivariate normal model, but it is not valid for every joint distribution. Use the model assumption before making the independence conclusion.
• Losing a sign in the density. Compute both standardized values before evaluating the cross-product. If one is negative and the other positive, their product is negative.
• Treating the endpoints as ordinary density cases. At \(\rho=\pm1\), the covariance matrix is singular, so the given two-dimensional density formula does not apply.